{n=2}}"},{"latex":"\\htmlData{anchor=squared}{\\color{@amber}{n^2=4}}"}],"highlights":[{"color":"sky","style":"text-color-with-background","anchor":"chosen"},{"color":"amber","style":"text-color-with-background","anchor":"squared"}],"connections":[{"to":"squared","from":"chosen","color":"sky","label":"square"}]},"order":1,"title":"Square the prime","description":"Now calculate the square of the chosen prime."},{"type":"text","order":2,"title":"State the disproof","content":"Since $4$ is even, this prime value of $n$ does **not** give an odd square. Therefore the word “always” is false, so the statement is disproved. Stating that $2^2=4$ is even earns **A1**.","highlights":[{"text":"$$n^2$ is always odd","location":"specification"}]}]}],"generatedAt":"2026-07-21T04:00:31.452Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]},{"id":"ef829299-ba74-4764-8df2-f13f0305d2a1","label":"b","content":"Find a counter example to disprove the statement: $(x + 2)^2$ is always greater than $x^2 + 2$","markscheme":"- Let $x = -2$. M1\n- LHS: $(-2+2)^2 = 0$. A1\n- RHS: $(-2)^2 + 2 = 6$. A1\n- $0$ is not $> 6$.","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Selects a value for $x$ (e.g., $x=-2$).","code":"M"},{"id":"p2","marks":1,"description":"Evaluates both sides and shows the statement is false.","code":"A"}],"finalAnswers":["x = -2"],"generatedAt":"2026-05-12T02:14:06.612Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"counterexample","label":"Choose a counterexample","steps":[{"type":"text","order":0,"title":"Choose a value that makes the bracket zero","content":"Choose $x=-2$, because it makes the bracket $x+2$ equal to zero. This is a valid counterexample value, and writing it down earns the M1 mark.","highlights":[{"text":"Find a counter example to disprove the statement","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\color{@sky}{\\left(-2+2\\right)^2}=0$"},{"latex":"$$\\color{@orange}{\\left(-2\\right)^2+2}=6$"},{"latex":"$$0\\not>6$"}]},"order":1,"title":"Evaluate both sides","description":"Substitute x = -2 into each side, then evaluate them separately. The left-hand-side calculation earns A1 and the right-hand-side calculation earns A1. Since 0 is not greater than 6, x = -2 disproves the ‘always’ statement."}]}],"generatedAt":"2026-07-21T04:03:36.072Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"07b9dc6a-1b72-4311-9c5e-072f0a0b629d","revisionId":"2b0c1f5c-f83f-4a7c-a92c-e295ee6a5c2d","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"easy","totalMarks":2,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"0e48f0c9-767b-4d7b-a958-ac8579a13f4b","label":null,"content":"The expression \\[(15x^2 + 4x + 9) / (5x - 2)\\] can be written as \\[Ax + B + \\frac{C}{5x - 2}\\] State the value of \$A\$.","markscheme":"- Sets up the identity $15x^2 + 4x + 9 = (Ax + B)(5x - 2) + C$ or performs the first step of algebraic division $(15x^2) \\div (5x)$ M1\n- Equates coefficients of $x^2$ to find $15 = 5A$ or completes the division to find $A = 3$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"polynomial_division","label":"Polynomial Long Division","steps":[{"type":"text","order":0,"title":"Set up long division","content":"To write the expression $\\frac{15x^2 + 4x + 9}{5x - 2}$ in the form $Ax + B + \\frac{C}{5x - 2}$, we can perform algebraic long division of the quadratic numerator by the linear denominator.\n\nWe start by setting up the division:\n\n$$5x - 2 \\ \\overline{) \\ 15x^2 + 4x + 9}$$"},{"type":"text","order":1,"title":"Divide leading terms","content":"Next, we divide the leading term of the dividend, $15x^2$, by the leading term of the divisor, $5x$.\n\n$$\\fra

Algebraic Methods

Question 15

Algebraic Methods Questions

Algebraic Methods

Question 15

Algebraic Methods Questions